I am interested in the unique solution to: $$x=(1-\frac{1+a}{n}+\frac{x(1+a)}{n})^n, (1)$$ such that $x\in(0,1)$. Specifically, I am interested in the case where $n\to\infty$.
I know that this is the extinction probability of $Bin[n,\frac{1+a}{n}]$-Galton-Watson tree, therefore I know that the solution should be $$1-\frac{2a-2a^2/3+O(a^4)}{1+a}. (2)$$ Furthermore, I know I can express $(1)$ by: $$x=e^{(x-1)(1+a)}. (3)$$.
And still, I don't understand - how do we get from $(3)$ to $(2)$? How do we approach $(3)$?
I am weakly familiar with the Lambert function - is this the idea here? How do we use it?
If you start with $(3)$ $$x=e^{(x-1)(1+a)}$$ the solution is given by $$x=-\frac 1{1+a}W\left(-(1+a)\, e^{-(1+a)}\right)$$ Expanding as series around $a=0$ (assuming $a>0$) $$W\left(-(1+a)\, e^{-(1+a)}\right)=-1+a-\frac{2 }{3}a^2+\frac{4 }{9}a^3-\frac{44 }{135}a^4+O\left(a^5\right)$$ making $$x=\frac{1-a+\frac{2 }{3}a^2-\frac{4 }{9}a^3+\frac{44 }{135}a^4+O\left(a^5\right) } {1+a}$$ or $$x=1-\frac{2 a-\frac{2 }{3}a^2+\frac{4 }{9}a^3-\frac{44 }{135}a^4+O\left(a^5\right) }{1+a}$$ wich is $(2)$.
Edit
This approximation is not very good (even very bad when $a>\frac 12$). You could have a much better one using a Padé approximant to obtain $$x=-\frac{2 b^3-18 b^2+45 b+16}{3 b\left(6 b^2+8 b+1\right)}\qquad \text{where} \qquad b=1+a$$ If you want much better ones, just tell me.
Details about the series expansion
For the expansion of $$W\left(-(1+a)\, e^{-(1+a)}\right)$$ around $a=0$, start with
$$-(1+a)\, e^{-(1+a)}=-\frac 1e+\frac 1e\sum_{n=0}^\infty(-1)^{n}\frac{ (n!-(n-1)!)}{ (n-1)! n!}a^n$$ that you can truncate whereever you want.
Now, around $t=-\frac1e$, $$W(t)=-1+x-\frac{x^2}{3}+\frac{11 x^3}{72}-\frac{43 x^4}{540}+\frac{769 x^5}{17280}-\frac{221 x^6}{8505}+O\left(x^7\right)$$ where $x=\sqrt{2(1+et)}$.
Combining both expansions
$$W\left(-(1+a)\, e^{-(1+a)}\right)=-1+a-\frac{2 a^2}{3}+\frac{4 a^3}{9}-\frac{44 a^4}{135}+\frac{104 a^5}{405}-\frac{40 a^6}{189}+O\left(a^7\right)$$